【USACO】Milking Cows

Three farmers rise at 5 am each morning and head for the barn to milk three cows. The first farmer begins milking his cow at time 300 (measured in seconds after 5 am) and ends at time 1000. The second farmer begins at time 700 and ends at time 1200. The third farmer begins at time 1500 and ends at time 2100. The longest continuous time during which at least one farmer was milking a cow was 900 seconds (from 300 to 1200). The longest time no milking was done, between the beginning and the ending of all milking, was 300 seconds (1500 minus 1200).

Your job is to write a program that will examine a list of beginning and ending times for N (1 <= N <= 5000) farmers milking N cows and compute (in seconds):

• The longest time interval at least one cow was milked.
• The longest time interval (after milking starts) during which no cows were being milked.

PROGRAM NAME: milk2

INPUT FORMAT

Line 1:	The single integer
Lines 2..N+1:	Two non-negative integers less than 1000000, the starting and ending time in seconds after 0500

SAMPLE INPUT (file milk2.in)

3
300 1000
700 1200
1500 2100

OUTPUT FORMAT

A single line with two integers that represent the longest continuous time of milking and the longest idle time.

SAMPLE OUTPUT (file milk2.out)

900 300

---

题解：简单的区间合并。先把时间区间按照开始时间排序，然后开始合并区间。区间之间的情况有以下几种：

黑色表示早开始的区间，红色表示在它之后开始的区间。那么第1，3中区间都不用合并，第2中情况需要将区间合并成一个长区间。然后用一个vector存放合并后的区间，最后再遍历vector，找到长度最长的区间，即为最长的持续时间，而区间之间的最长时间即为最长闲置时间。代码如下：

/*ID:Moment1991
  PROG:milk2
  LANG:C++
  Compiling...
  Compile: OK

  Executing...
   Test 1: TEST OK [0.003 secs, 3496 KB]
   Test 2: TEST OK [0.000 secs, 3496 KB]
   Test 3: TEST OK [0.000 secs, 3496 KB]
   Test 4: TEST OK [0.008 secs, 3496 KB]
   Test 5: TEST OK [0.003 secs, 3496 KB]
   Test 6: TEST OK [0.005 secs, 3496 KB]
   Test 7: TEST OK [0.008 secs, 3496 KB]
   Test 8: TEST OK [0.019 secs, 3496 KB]

  All tests OK.
 */
#include <iostream>
#include <stdlib.h>
#include <vector>
#include <fstream>
#include <vector>
using namespace std;
struct interval{
    int start;
    int end;
};
int cmp( const void *a ,const void *b)
{
    return (*(interval *)a).start > (*(interval *)b).start ?  : -;
}
int main(){
    ifstream cin("milk2.in");
    ofstream cout("milk2.out");

    struct interval INTERVAL[];
    int n;

    //读入输入
    cin >> n;
    for(int i = ;i < n;i ++){
        cin >> INTERVAL[i].start>>INTERVAL[i].end;
    }
    //根据开始时间对区间排序
    qsort(INTERVAL,n,sizeof(INTERVAL[]),cmp);

    int max_busy = ;
    int max_free = ;

    struct interval temp;    temp.start = INTERVAL[].start;
    temp.end = INTERVAL[].end;

    //合并区间，将合并后的区间存放在vector AFTER_MERGE中
    vector <interval> AFTER_MERGE;
    for(int i = ;i < n;i++){
        temp.start = INTERVAL[i].start;
        temp.end = INTERVAL[i].end;

        while(i+<n && INTERVAL[i+].start <= temp.end){
            if(INTERVAL[i+].end > temp.end)
                temp.end =  INTERVAL[i+].end;
            i++;
        }
        AFTER_MERGE.push_back(temp);
    }

    //遍历AFTER_MERGE，找到最长的区间以及区间之间最长的闲置时间
    for(int i = ;i < AFTER_MERGE.size();i++)
    {
        if(AFTER_MERGE[i].end - AFTER_MERGE[i].start > max_busy)
            max_busy= AFTER_MERGE[i].end - AFTER_MERGE[i].start;

        if(i+<AFTER_MERGE.size() && AFTER_MERGE[i+].start - AFTER_MERGE[i].end > max_free)
            max_free = AFTER_MERGE[i+].start - AFTER_MERGE[i].end;
    }
    cout << max_busy<<" "<<max_free<<endl;
}